Supremum is Unique

Theorem

Let $\struct {S, \preceq}$ be an ordered set.

Let $T$ be a non-empty subset of $S$.

Then $T$ has at most one supremum in $S$.


Proof

Let $c$ and $c'$ both be suprema of $T$ in $S$.

From the definition of supremum, $c$ and $c'$ are upper bounds of $T$ in $S$.


By that definition:

$c$ is an upper bound of $T$ in $S$ and $c'$ is a supremum of $T$ in $S$ implies that $c' \preceq c$
$c'$ is an upper bound of $T$ in $S$ and $c$ is a supremum of $T$ in $S$ implies that $c \preceq c'$.


So:

$c' \preceq c \land c \preceq c'$

and thus by the antisymmetry of the ordering $\preceq$:

$c = c'$

$\blacksquare$


Sources

  • 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.3$
  • 1967: Michael Spivak: Calculus ... (previous) ... (next): Part $\text {II}$: Foundations: Chapter $8$: Least Upper Bounds
  • 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $1$: Review of some real analysis: $\S 1.1$: Real Numbers
  • 1982: Peter T. Johnstone: Stone Spaces ... (previous) ... (next): Chapter $\text I$: Preliminaries, Definition $1.2$
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